Region $R$ is enclosed by the curve $y=2\sqrt{x-3}$, the line $y=4$, and the line $x=3$. $y$ $x$ ${y=2\sqrt{x-3}}$ ${y=4}$ $ 3$ $ 7$ $ R$ What is the volume of the solid generated when $R$ is rotated about the $y$ -axis? Give an exact answer in terms of $\pi$.
Answer: Let's imagine the solid is made out of many thin slices. Each slice is a cylinder with a hole in the middle, much like a washer. $y$ $x$ ${y=2\sqrt{x-3}}$ ${y=4}$ $ 3$ $ 7$ Let the width of each slice be $dy$, let the radius of the washer, as a function of $y$, be $r_1(y)$, and let the radius of the hole, as a function of $y$, be $r_2(y)$. Then, the volume of each slice is $\pi[(r_1(y))^2-(r_2(y))^2]\,dy$, and we can sum the volumes of infinitely many such slices with an infinitely small width using a definite integral: $\int_a^b \pi [(r_1(y))^2-(r_2(y))^2]\,dy$ We call this the washer method. What we now need is to figure out the expressions for $r_1(y)$ and $r_2(y)$ and the interval of integration. $r_1(y)$ is equal to the distance from curve $y=2\sqrt{x-3}$ to the $y$ -axis. To find it, we need to solve the equation for $x$ : $x=\dfrac14 y^2+3$ So, $r_1(y)=\dfrac14 y^2+3}$. $r_2(y)$ is equal to the distance from the line $x=3$ to the $y$ -axis. So, ${r_2(y)=3}$. Now we can find an expression for the area of the washer's base: $\begin{aligned} &\phantom{=} \pi [(r_1(y)})^2-({r_2(y)})^2] \\\\ &= \pi \left[ \left( {\dfrac14 y^2+3}\right)^2-({3})^2 \right] \\\\ &=\pi\left( \dfrac{1}{16}y^4+\dfrac32y^2+9-9 \right) \\\\ &=\pi \left( \dfrac{1}{16}y^4+\dfrac32y^2 \right) \end{aligned}$ The bottom endpoint of $R$ is at $y=0$ and the top endpoint is at $y=4$. So the interval of integration is $[0,4]$. Now we can express the definite integral in its entirety! $\int_0^4 \left[\pi \left( \dfrac{1}{16}y^4+\dfrac32y^2 \right) \right]dy$ Let's evaluate the integral. $\int_0^4 \left[\pi\ \left( \dfrac{1}{16}y^4+\dfrac32y^2 \right) \right]dy=\dfrac{224\pi}{5}$ In conclusion, the volume of the solid is $\dfrac{224\pi}{5}$.